Show that the tangents to the curves y=e^x, y=e^-x

Show that the tangents to the curves $y=e^{x}$, $y=e^{-x}$, and $y=\tfrac12(e^{x}+e^{-x})$ drawn at points with the same abscissa $x=a$ have a common point.

Step 1: Tangent to $y=e^{x}$ at $x=a$

Slope $k_{1}=e^{a}$, point $(a,e^{a})$: \[\ell_{1}\colon\ y = e^{a}(x-a+1). \tag{1}\]

Step 2: Tangent to $y=e^{-x}$ at $x=a$

Slope $k_{2}=-e^{-a}$, point $(a,e^{-a})$: \[\ell_{2}\colon\ y = e^{-a}(1+a-x). \tag{2}\]

Step 3: Tangent to $y=\tfrac12(e^{x}+e^{-x})$ at $x=a$

Slope $k_{3}=\tfrac12(e^{a}-e^{-a})$, point $\left(a,\tfrac{e^a+e^{-a}}2\right)$: \[\ell_{3}\colon\ y = \frac{e^{a}-e^{-a}}{2}\,x + \frac{e^{a}+e^{-a}-a(e^{a}-e^{-a})}{2}. \tag{3}\]

Step 4: Intersection of $\ell_{1}$ and $\ell_{2}$

Setting (1) equal to (2) and writing $u=x-a$: $e^{a}(u+1)=e^{-a}(1-u)$, so $u(e^{a}+e^{-a})=e^{-a}-e^{a}$, i.e.

\[ u = -\tanh a, \qquad x^{*} = a-\tanh a, \qquad y^{*} = e^{a}(u+1) = \frac{2}{e^{a}+e^{-a}}. \tag{4} \]

Step 5: $S=(x^{*},y^{*})$ lies on $\ell_{3}$ too

With $P=e^{a}$, $Q=e^{-a}$ (so $PQ=1$), substituting $x^{*}$ into (3) and combining over $2(P+Q)$, the numerator becomes $-(P-Q)^{2}+(P+Q)^{2}=4PQ=4$: \[\ell_{3}(x^{*}) = \frac{-(P-Q)^{2}+(P+Q)^{2}}{2(P+Q)} = \frac{4}{2(P+Q)} = \frac{2}{e^{a}+e^{-a}} = y^{*}. \tag{5}\]

Step 6: Verification

\[a=1.2\text{: } \ell_1(x^*)-y^*=0,\ \ell_2(x^*)-y^*=0,\ \ell_3(x^*)-y^*=0 \text{ (numerically)}\ \checkmark\]